LOJ#6281. 数列分块入门 5

数据最大就2^31次方，通过计算这些数开方五次就变成1了，然后把继续开方也不会变了，所以对于每个块，前5次更新暴力，之后就可以直接跳过，查询就直接就可以了

#include<map>
#include<set>
#include<ctime>
#include<cmath>
#include<stack>
#include<queue>
#include<string>
#include<vector>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define first fi
#define second se
#define lowbit(x) (x & (-x))

typedef unsigned long long int ull;
typedef long long int ll;
const double pi = 4.0*atan(1.0);
const int inf = 0x3f3f3f3f;
const int maxn = 50005;
const int maxm = 100000;
const int mod = 998244353;
using namespace std;

int n, m, tol, T;
int block;
int a[maxn];
int b[maxn];
int c[maxn];
int belong[maxn];

void init() {
    memset(a, 0, sizeof a);
    memset(b, 0, sizeof b);
    memset(c, 0, sizeof c);
    memset(belong, 0, sizeof belong);
}

int L(int x) {
    return (x-1) * block + 1;
}

int R(int x) {
    return min(n, x*block);
}

void update(int l, int r) {
    for(int i=l; i<=min(r, R(belong[l])); i++) {
        b[belong[i]] -= a[i];
        a[i] = sqrt(a[i]);
        b[belong[i]] += a[i];
    }
    if(belong[l] == belong[r])    return ;
    if(belong[l] != belong[r]) {
        for(int i=L(belong[r]); i<=r; i++) {
            b[belong[i]] -= a[i];
            a[i] = sqrt(a[i]);
            b[belong[i]] += a[i];
        }        
    }
    for(int i=belong[l]+1; i<belong[r]; i++) {
        if(c[i] == 6)    continue;
        for(int j=L(i); j<=R(i); j++) {
            b[i] -= a[j];
            a[j] = sqrt(a[j]);
            b[i] += a[j];
        }
        c[i]++ ;
    }
}

int query(int l, int r) {
    int ans = 0;
    for(int i=l; i<=min(r, R(belong[l])); i++)    ans += a[i];
    if(belong[l] == belong[r])    return ans;
    if(belong[l] != belong[r]) {
        for(int i=L(belong[r]); i<=r; i++) {
            ans += a[i];
        }
    }
    for(int i=belong[l]+1; i<belong[r]; i++)    ans += b[i];
    return ans;
}

int main() {
    while(~scanf("%d", &n)) {
        block = sqrt(n);
        for(int i=1; i<=n; i++) {
            scanf("%d", &a[i]);
            belong[i] = (i-1) / block + 1;
            b[belong[i]] += a[i];
        }
        m = n;
        while(m--) {
            int op, l, r, c;
            scanf("%d%d%d%d", &op, &l, &r, &c);
            if(op == 0) {
                update(l, r);
            } else {
                int ans = query(l, r);
                printf("%d
", ans);
            }
        }
    }
    return 0;
}